For a system of simultaneous linear equations | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 &

Vedclass · Accepted Answer

$\left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$

Vedclass · Answer

(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight]$.We know that $A^{-1} = \frac{1}{|A|} \operatorname{Adj} A$.Multiplying $A X = B$ by $A^{-1}$ on both sides, we get $X = A^{-1} B = \frac{1}{|A|} \operatorname{Adj} A \cdot B$.$X = \frac{1}{|A|} \left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$.Performing matrix multiplication:$X = \frac{1}{|A|} \left[\begin{array}{c}1(1) + (-1)(1) + (-1)(2) \ 1(1) + 1(1) + (-1)(2) \ 1(1) + 1(1) + 1(2)\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}1 - 1 - 2 \ 1 + 1 - 2 \ 1 + 1 + 2\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight]$.Since $|A| > 0$, we check the determinant of the given $\operatorname{Adj} A$. We know $|\operatorname{Adj} A| = |A|^{n-1}$, where $n=3$.$|\operatorname{Adj} A| = 1(1+1) - (-1)(1+1) + (-1)(1-1) = 2 + 2 + 0 = 4$.So, $|A|^2 = 4 \implies |A| = 2$ (since $|A| > 0$).Substituting $|A| = 2$ into the expression for $X$:$X = \frac{1}{2} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight] = \left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$.Thus, option $A$ is correct.

For a system of simultaneous linear equations, if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$, $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$, then $X=$

Similar Questions

Let $\lambda$ be a real number for which the system of linear equations $x + y + z = 6$,$4x + \lambda y - \lambda z = \lambda - 2$,and $3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:

Find the matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

Solve the system of linear equations using the matrix method: $x-y+2z=7$,$3x+4y-5z=-5$,$2x-y+3z=12$.

If the homogeneous system of linear equations $x-2y+3z=0, 2x+4y-5z=0, 3x+\lambda y+\mu z=0$ has a non-trivial solution, then $8\mu+11\lambda=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module